Question

a constant force acts on an object of mass 5 kg for a duration of 2 seconds. it increases the velocity from 3m/s to 7m/s. Find the magnitude of applied force. Now, if force was applied or the duration of 10 sec, what would be the final velocity?​

Answer 1

Answer:

F = 10N

If t = 10s

Then, v = -17m/s

Explanation:

m = 5kg

t = 2s

u = 3m/s

v = 7m/s

F = ma

F = 5kg * a

[a = (v - u) /t]

F = 5kg * (3 - 7)/2

F = 5kg * -4/2m/s^2

F = 5kg * -2m/s^2

F = -10N

If t = 10s

F = m * (v - u/t)

-10N = 5kg * (v - 3/10)

-10/5 = (v - 3/10)

-2 * 10 = v - 3

-20 = v - 3

-v = 20 - 3

-v = 17

v = -17m/s

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